Matrix/2x2/Shearing matrices/Eigenvalues and diagonalizable/Example

We consider ${}2\times 2$-shearing matrices

${\begin{pmatrix}1&a\\0&1\end{pmatrix}}$

with ${}a\in K$. The condition for some ${}\lambda \in K$ to be an eigenvalue means

${}{\begin{pmatrix}1&a\\0&1\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,.$

This yields the equations

$x+ay=\lambda x{\text{ and }}y=\lambda y.$

For ${}\lambda \neq 1$, we get ${}y=0$ and hence also ${}x=0$, that is, only ${}1$ can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes

$x+ay=x{\text{ and }}ay=0.$

For ${}a\neq 0$, we get ${}y=0$ and thus ${}{\begin{pmatrix}x\\0\end{pmatrix}}$ is the eigenspace for the eigenvalue ${}1$, and ${}{\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector which spans this eigenspace. For ${}a=0$, we have the identity matrix, and the eigenspace for the eigenvalue ${}1$ is the total plane. For ${}a\neq 0$, there is a one-dimensional eigenspace, and the mapping is not diagonalizable.